1. Given function

$$f(x)=\int_{1/x}^{x} \frac{e^{-(t+1/t)}}{t}\,dt, \qquad x>0$$

Let

$$g(t)=\frac{e^{-(t+1/t)}}{t}, \qquad t>0.$$

Then

$$f(x)=\int_{1/x}^{x} g(t)\,dt.$$

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2. Check option C: relation between $f(x)$ and $f(1/x)$

We compute:

$$f\left(\frac{1}{x}\right)=\int_{x}^{1/x} g(t)\,dt.$$

Reversing limits,

$$f\left(\frac{1}{x}\right)=-\int_{1/x}^{x} g(t)\,dt=-f(x).$$

Hence,

$$f(x)+f\left(\frac{1}{x}\right)=0 \qquad \forall x>0.$$

So Option C is correct.

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3. Differentiate $f(x)$ using Leibniz rule

Since

$$f(x)=\int_{1/x}^{x} g(t)\,dt,$$

we get

$$f'(x)=g(x)\cdot \frac{d}{dx}(x)-g(1/x)\cdot \frac{d}{dx}(1/x).$$

Now,

$$\frac{d}{dx}\left(\frac1x\right)=-\frac1{x^2}.$$

So

$$f'(x)=g(x)+\frac{g(1/x)}{x^2}.$$

Now evaluate $g(1/x)$:

$$g\left(\frac1x\right)=\frac{e^{-\left(\frac1x+x\right)}}{1/x}=x e^{-(x+1/x)}.$$

Thus,

$$\frac{g(1/x)}{x^2}=\frac{x e^{-(x+1/x)}}{x^2}=\frac{e^{-(x+1/x)}}{x}.$$

Also,

$$g(x)=\frac{e^{-(x+1/x)}}{x}.$$

Therefore,

$$f'(x)=\frac{e^{-(x+1/x)}}{x}+\frac{e^{-(x+1/x)}}{x} =\frac{2e^{-(x+1/x)}}{x}.$$

Since $x>0$ and exponential is always positive,

$$f'(x)>0 \qquad \forall x>0.$$

So $f$ is strictly increasing on $(0,\infty)$.

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4. Check option A

Option A says: $f(x)$ is monotonically increasing on $[1,\infty)$.

Since $f'(x)>0$ for all $x>0$, this is true.

So Option A is correct.

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5. Check option B

Option B says: $f(x)$ is monotonically decreasing on $(0,1)$.

But we found

$$f'(x)>0 \qquad \forall x>0,$$

so $f$ is increasing on $(0,1)$ as well, not decreasing.

Therefore Option B is false.

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6. Check option D: oddness of $f(2^x)$

Define

$$h(x)=f(2^x).$$

Then

$$h(-x)=f(2^{-x})=f\left(\frac{1}{2^x}\right).$$

Using option C result,

$$f\left(\frac{1}{2^x}\right)=-f(2^x).$$

Hence,

$$h(-x)=-h(x).$$

So $h(x)=f(2^x)$ is an odd function on $\mathbb{R}$.

Therefore Option D is correct.

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7. Final conclusion

Correct options are:

$$\boxed{A,\ C,\ D}$$

This matches the stored correct answer.